Evaluate ∫ d x x ( 1 + x ) .

I = ∫ d x x ( 1 + x ) . Let t = 1 + x . Differentiate t , with respect to x . ⇒ d ( t ) d x = d 1 + x d x . ⇒ d t d x = 1 2 x , using the basic of differentiation. ⇒ 2 d t = d x x . Now I = ∫ 2 d t t . ⇒ I = ∫ 2 t - 1 2 d t . ⇒ I = 2 t - 1 2 + 1 - 1 2 + 1 + c , using ∫ x n d x = x n + 1 n + 1 + c . ⇒ I = 4 t 1 2 1 + c , where c , is a constant. ⇒ I = 4 ( 1 + x ) 1 2 + c .

Evaluate ∫ d x x ( 1 + x ) .

I = ∫ d x x ( 1 + x ) . Let t = 1 + x . Differentiate t , with respect to x . ⇒ d ( t ) d x = d 1 + x d x . ⇒ d t d x = 1 2 x , using the basic of differentiation. ⇒ 2 d t = d x x . Now I = ∫ 2 d t t . ⇒ I = ∫ 2 t - 1 2 d t . ⇒ I = 2 t - 1 2 + 1 - 1 2 + 1 + c , using ∫ x n d x = x n + 1 n + 1 + c . ⇒ I = 4 t 1 2 1 + c , where c , is a constant. ⇒ I = 4 ( 1 + x ) 1 2 + c .

Evaluate ∫ x 2 + a 2 x d x

Let I = ∫ x 2 + a 2 x d x Put x = a tan θ ⇒ d x = a sec 2 θ d θ I = ∫ a 2 tan 2 θ + a 2 a tan θ × a sec 2 θ d θ ⇒ I = a ∫ sec 3 θ tan θ d θ ⇒ I = a ∫ sec 2 θ sin θ d θ ⇒ I = a ∫ 1 + tan 2 θ sin θ d θ ⇒ I = a ∫ cosec θ + sin 2 θ cos 2 θ sin θ d θ ⇒ I = a ∫ cosec θ d θ + a ∫ sin θ cos 2 θ d θ ⇒ I = a ∫ cosec θ d θ - a ∫ d cos θ cos 2 θ ⇒ I = a log cosec θ - cot θ + a sec θ + C

Evaluate ∫ e tan - 1 x 1 + x 2 d x

Let I = ∫ e tan - 1 x 1 + x 2 d x Let tan - 1 x = t ⇒ 1 1 + x 2 d x = d t ∴ I = ∫ e t d t = e t + C = e tan - 1 x + C .

Evaluate ∫ sin x + cos x sin 2 x d x

To evaluate ∫ sin x + cos x sin 2 x d x , substitute sin x - cos x = t ⇒ cos x + sin x d x = d t and also sin x - cos x 2 = t 2 ⇒ sin 2 x + cos 2 x - 2 sin x cos x = t 2 ⇒ sin 2 x = 1 - t 2 ∴ ∫ sin x + cos x sin 2 x d x = ∫ 1 1 - t 2 d t = sin - 1 t + C = sin - 1 sin x - cos x + C .

Evaluate ∫ d x 5 - 4 x - 2 x 2

Let I = ∫ d x 5 - 4 x - 2 x 2 I = ∫ d x - 2 x 2 + 2 x - 5 2 = 1 2 ∫ d x - x 2 + 2 x + 1 2 - 1 2 - 5 2 = 1 2 ∫ d x 7 2 - x + 1 2 = 1 2 ∫ d x 7 2 2 - x + 1 2 = 1 2 sin - 1 x + 1 7 2 + c I = 1 2 sin - 1 2 x + 1 7 + c

Evaluate ∫ ( x - 4 ) e x ( x - 2 ) 3 d x .

I = ∫ ( x - 4 ) e x ( x - 2 ) 3 d x . ⇒ I = ∫ ( x - 2 - 2 ) e x ( x - 2 ) 3 d x . ⇒ I = ∫ ( x - 2 ) ( x - 2 ) 3 - 2 x - 2 3 e x d x . ⇒ I = ∫ 1 ( x - 2 ) 2 - 2 x - 2 3 e x d x Let f ( x ) = 1 x - 2 2 . Differentiating f ( x ) , with respect to x . d ( f ( x ) d x = d 1 x - 2 2 d x ⇒ f ' ( x ) = - 2 x - 2 3 ⇒ I = ∫ f ( x ) + f ' ( x ) e x d x ⇒ I = f ( x ) . e x + c ⇒ I = 1 x - 2 2 . e x + c , where c , is a constant.

Evaluate ∫ x 2 + x + 1 ( x + 2 ) x 2 + 1 d x .

I = ∫ x 2 + x + 1 ( x + 2 ) x 2 + 1 d x . As we can see that the integral in the form of ∫ Q L . Q 1 d y , then the Q , can be substituted with A L + B x + C Q 1 . . . ˙ x 2 + x + 1 ( x + 2 ) x 2 + 1 = A ( x + 2 ) + B x + C x 2 + 1 ⇒ . . ˙ x 2 + x + 1 ( x + 2 ) x 2 + 1 = A ( x 2 + 1 ) + B x + C ( x + 2 ) ( x + 2 ) ( x 2 + 1 ) On comparing the coefficients on both side of the equation. ⇒ A + B = 1 & 2 B + C = 1 & A + 2 C = 1 ⇒ A = 3 5 , B = 2 5 & C = 1 5 . . . ˙ I = ∫ 3 5 x + 2 + 2 5 x + 1 5 x 2 + 1 d x ⇒ I = 3 5 ∫ d x x + 2 + 2 5 ∫ x d x x 2 + 1 + 1 5 ∫ d x x 2 + 1 . ⇒ I = 3 5 ln x + 2 + c 1 + 2 5 ∫ x d x x 2 + 1 + 1 5 tan - 1 x 1 + c 2 , using d y y = ln y + c & ∫ d x x 2 + a 2 = 1 a tan - 1 x a + c , where c , c 1 , c 2 , are integration constant. Let t = x 2 + 1 Differentiate t , with respect to x . d t d x = d ( x 2 + 1 ) d x d t d x = 2 x , using d ( y n ) d y = n y n - 1 & d ( C ) d y = 0 , where C , is a constant. d t = 2 x d x . ⇒ I = 3 5 ln x + 2 + c 1 + 1 5 ∫ d t t + 1 5 tan - 1 x + c 2 ⇒ I = 3 5 ln x + 2 + c 1 + 1 5 ln t + c 3 + 1 5 tan - 1 x + c 2 . ⇒ I = 3 5 ln x + 2 + 1 5 ln x 2 + 1 + 1 5 tan - 1 x + C , where C , is an integration constant.

Evaluate ∫ tan x + tan 3 x 1 + tan 3 x d x .

I = ∫ tan x + tan 3 x 1 + tan 3 x d x . ⇒ I = ∫ tan x ( 1 + tan 2 x ) 1 + tan 3 x d x . ⇒ I = ∫ tan x × sec 2 x 1 + tan 3 x d x . Let t = tan x . Differentiating t , with respect to x . ⇒ d t d x = d ( tan x ) d x . ⇒ d t d x = sec 2 x ⇒ d t = sec 2 x d x ⇒ I = ∫ t d t 1 + t 3 Adding and subtracting in the numerator by t 2 . ⇒ I = ∫ t 2 - t 2 + t 1 + t 3 d t . ⇒ I = ∫ t 2 + t 1 + t 3 d t - ∫ t 2 1 + t 3 d t . ⇒ I = ∫ t ( 1 + t ) ( 1 + t ) ( 1 + t 2 - t ) d t - ∫ t 2 1 + t 3 d t , using ( a 3 + b 3 ) = ( a + b ) ( a 2 + b 2 - a b ) . ⇒ I = ∫ t ( 1 + t 2 - t ) d t - ∫ t 2 1 + t 3 d t . Let 1 + t 3 = u . Differentiate u , with respect to t d u d t = d ( 1 + t 3 ) d t . d u 3 = t 2 d t . ⇒ I = ∫ t ( 1 + t 2 - t ) d t - 1 3 ∫ d u u . ⇒ I = ∫ t ( 1 + t 2 - t ) d t - 1 3 ln u + c 1 , using d x x = ln x + C ' , where C ' , is an integration constant. ⇒ I = ∫ t ( 1 + t 2 - t ) d t - 1 3 ln 1 + tan 3 x + c 1 . Now dividing and multiplying the numerator of first integration by 2 . ⇒ I = 1 2 ∫ 2 t ( 1 + t 2 - t ) d t - ln 1 + tan 3 x + c 1 . ⇒ I = 1 2 ∫ 2 t - 1 + 1 ( 1 + t 2 - t ) d t - ln 1 + tan 3 x + c 1 . ⇒ ⇒ I = 1 2 ∫ 2 t - 1 2 ( 1 + t 2 - t ) d t + 1 2 ∫ 1 ( 1 + t 2 - t ) d t - 1 3 ln 1 + tan 3 x + c 1 . Let 1 + t 2 - t = v , from the first integration of I . Differentiating v , with respect to t . ⇒ d v d t = d ( 1 + t 2 - t ) d t ⇒ d v = 2 t - 1 d t . . . ˙ ⇒ I = 1 2 ∫ d v v + 1 2 ∫ 1 ( 1 + t 2 - t ) d t - 1 3 ln 1 + tan 3 x + c 1 . . . ˙ ⇒ I = 1 2 ln v + c 2 + 1 2 ∫ 1 t - 1 2 2 + 3 2 2 d t - 1 3 ln 1 + tan 3 x + c 1 , using completing square method in the denominator. ⇒ I = 1 2 ln t 2 - t + 1 + c 2 + 1 2 × 1 3 2 tan - 1 t - 1 2 3 2 + c 3 - 1 3 ln 1 + tan 3 x + c 1 , using 1 x 2 + a 2 = 1 a tan - 1 x a + C , where C , is an integration constant. Here c 1 , c 2 & c 3 , are the integration constant. ⇒ I = 1 2 ln t 2 - t + 1 + 3 tan - 1 2 t - 1 2 3 - 1 3 ln 1 + tan 3 x + c . ⇒ I = 1 2 ln t 2 - t + 1 + 3 tan - 1 2 t - 1 3 - 1 3 ln 1 + tan 3 x + c . ⇒ I = 1 2 ln tan 2 x - tan x + 1 + 1 3 tan - 1 2 tan x - 1 3 - 1 3 ln 1 + tan 3 x + c .

Evaluate ∫ x + 3 x 2 - 2 x - 5 d x

Let I = ∫ x + 3 x 2 - 2 x - 5 d x I = ∫ 1 2 ( 2 x - 2 ) + 4 x 2 - 2 x - 5 I = 1 2 ∫ 2 x - 2 x 2 - 2 x - 5 d x + 4 ∫ 1 x 2 - 2 x - 5 d x let x 2 - 2 x + 5 = t ⇒ ( 2 x - 2 ) d x = d t I = 1 2 ∫ d t t + 4 ∫ 1 x 2 - 2 x + 1 2 - 1 2 - 5 I = 1 2 log | t | + 4 ∫ 1 x - 1 2 - 6 d x + c 1 I = 1 2 log | x 2 - 2 x + 5 | + 4 ∫ 1 x - 1 2 - 6 2 d x + c 1 I = 1 2 log | x 2 - 2 x + 15 | + 4 × 1 2 6 log | x - 1 - 6 x - 1 + 6 | + c I = 1 2 log | x 2 - 2 x + 15 | + 2 6 log | x - 1 - 6 x - 1 + 6 | + c

Evaluate ∫ d x 5 + 4 cos x .

Given that, We have to evaluate ∫ d x 5 + 4 cos x . ⇒ ∫ d x 5 + 4 cos x = ∫ d x 5 + 4 1 - tan 2 x 2 1 + tan 2 x 2 ∵ cos θ = 1 - tan 2 θ 2 1 + tan 2 θ 2 ⇒ ∫ d x 5 + 4 cos x = ∫ 1 + tan 2 x d x 5 1 + tan 2 x 2 + 4 1 - tan 2 x 2 ⇒ ∫ d x 5 + 4 cos x = ∫ s e c 2 x d x 5 1 + tan 2 x 2 + 4 1 - tan 2 x 2 ∵ 1 + tan 2 θ = s e c 2 θ ⇒ ∫ d x 5 + 4 cos x = ∫ s e c 2 x d x 9 + tan 2 x 2 Put tan x 2 = t ⇒ 1 2 s e c 2 x 2 d x = d t ⇒ ∫ s e c 2 x d x 9 + tan 2 x 2 = ∫ 2 d t 9 + t 2 = 2 ∫ d t 3 2 + t 2 = 2 3 tan - 1 t 3 + C ∵ ∫ d x a 2 + x 2 = 1 a tan - 1 x a Put t = tan x 2 , ⇒ 2 3 tan - 1 tan x 2 3 + C Therefore, ∫ d x 5 + 4 cos x = 2 3 tan - 1 tan x 2 3 + C .

Evaluate ∫ 2 x - 1 ( x - 1 ) ( x + 2 ) ( x - 3 ) d x

Given, ∫ 2 x - 1 ( x - 1 ) ( x + 2 ) ( x - 3 ) d x Since the denominator of the integrand contains three non-repeated linear factors. So, we can write the integrand into the sum of three partial fractions. Thus, we have 2 x - 1 ( x - 1 ) ( x + 2 ) ( x - 3 ) = A x - 1 + B x + 2 + C x - 3 . . . . i , where A , B , and C are real numbers. ⇒ 2 x - 1 ( x - 1 ) ( x + 2 ) ( x - 3 ) = A x + 2 x - 3 + B x - 1 x - 3 + C x - 1 x + 2 ( x - 1 ) ( x + 2 ) ( x - 3 ) ⇒ 2 x - 1 = A x + 2 x - 3 + B x - 1 x - 3 + C x - 1 x + 2 . . . . i i Substituting x = 3 in equation i i , we have C 2 5 = 2 3 - 1 = 5 ⇒ C = 1 2 Now, Substituting x = - 2 in equation i i , we have 2 - 2 - 1 = B - 3 - 5 ⇒ - 4 - 1 = 15 B ⇒ B = - 1 3 Now, Substituting x = 1 in equation i i , we have ⇒ 2 1 - 1 = A 1 + 2 1 - 3 ⇒ 1 = - 6 A ⇒ A = - 1 6 Thus, from equation i , we have 2 x - 1 ( x - 1 ) ( x + 2 ) ( x - 3 ) = - 1 6 x - 1 - 1 3 x + 2 + 1 2 x - 3 Thus, ∫ 2 x - 1 ( x - 1 ) ( x + 2 ) ( x - 3 ) d x = ∫ - 1 6 x - 1 - 1 3 x + 2 + 1 2 x - 3 d x = - 1 6 ∫ 1 x - 1 d x - 1 3 ∫ 1 x + 2 d x + 1 2 ∫ 1 x - 3 d x = - 1 6 log x - 1 - 1 3 log x + 2 + 1 2 log x - 3 + C , where C is an integral constant. Hence, ∫ 2 x - 1 ( x - 1 ) ( x + 2 ) ( x - 3 ) d x = = - 1 6 log x - 1 - 1 3 log x + 2 + 1 2 log x - 3 + C , where C is an integral constant.